$ \newcommand{ \F }{ \mathcal{F} } \newcommand{ \prob }{ \mathbb{P} } \newcommand{ \G }{ \mathcal{G} } \newcommand{ \L }{ \mathcal{L} } \newcommand{ \exp }[1]{ \mathbb{E}\left[ #1 \right] } \newcommand{ \ind }[1]{ 1_{ #1 } } $
$\exp{X|Y}=Y \text{ almost surely}$ and $\exp{Y|X}=X \text{ almost surely}.$
Show that $P(X=Y)=1$.
Attempt
I know that if $X \in \L^1(\Omega, \F,\prob)$, $\G \subset \F$ then there exists $Y$ that is (i) $\G$ measurable, (ii) $\exp{|Y|} < \infty $ (iii) $\exp{ Y \ind{ F} }= \exp{ X \ind{G } } $ for all $G \in \G$.
\begin{align} P(X=Y)=1 &\Leftrightarrow P(X \neq Y ) = 0 \\ &\Leftrightarrow P( \{X<Y\}\cup\{Y<X \}) =0\\ &\Leftrightarrow P(X<Y) + P(X>Y) =0 \\ &\Leftrightarrow P(X<Y) = P(X>Y) =0 \\ \end{align}
I am not sure where to go from here.
